Triangle Properties Involving Half Angle Tangents and Side Lengths

To determine which of the given statements are true, we start with the given trigonometric identity involving the half-angles of $\Delta XYZ$:

$\tan\frac{X}{2} + \tan\frac{Z}{2} = \frac{2y}{x + y + z}$

Let $s = \frac{x + y + z}{2}$ denote the semi-perimeter of the triangle. The right-hand side of the equation can be rewritten as:

$\frac{2y}{x + y + z} = \frac{2y}{2s} = \frac{y}{s}$

Using the standard half-angle formulas in terms of semi-perimeter $s$:

$\tan\frac{X}{2} = \sqrt{\frac{(s-y)(s-z)}{s(s-x)}} \quad \text{and} \quad \tan\frac{Z}{2} = \sqrt{\frac{(s-x)(s-y)}{s(s-z)}}$

Adding these two expressions together gives:

$\tan\frac{X}{2} + \tan\frac{Z}{2} = \sqrt{\frac{s-y}{s}} \left( \sqrt{\frac{s-z}{s-x}} + \sqrt{\frac{s-x}{s-z}} \right)$

$\tan\frac{X}{2} + \tan\frac{Z}{2} = \sqrt{\frac{s-y}{s}} \left( \frac{(s-z) + (s-x)}{\sqrt{(s-x)(s-z)}} \right)$

Notice that $(s-z) + (s-x) = 2s - x - z = (x + y + z) - x - z = y$. Substituting this into the numerator yields:

$\tan\frac{X}{2} + \tan\frac{Z}{2} = \sqrt{\frac{s-y}{s}} \cdot \frac{y}{\sqrt{(s-x)(s-z)}}$

Equating this to $\frac{y}{s}$:

$\sqrt{\frac{s-y}{s}} \cdot \frac{y}{\sqrt{(s-x)(s-z)}} = \frac{y}{s}$

Dividing both sides by $y$ and multiplying by $\sqrt{s}$:

$\frac{\sqrt{s-y}}{\sqrt{(s-x)(s-z)}} = \frac{1}{\sqrt{s}}$

Squaring both sides and cross-multiplying:

$s(s-y) = (s-x)(s-z)$

Now, substitute $s = \frac{x + y + z}{2}$ back into the equation:

$\left(\frac{x + y + z}{2}\right)\left(\frac{x - y + z}{2}\right) = \left(\frac{-x + y + z}{2}\right)\left(\frac{x + y - z}{2}\right)$

$\frac{(x + z)^2 - y^2}{4} = \frac{y^2 - (x - z)^2}{4}$

$(x + z)^2 - y^2 = y^2 - (x - z)^2$

$(x + z)^2 + (x - z)^2 = 2y^2$

$2(x^2 + z^2) = 2y^2 \implies x^2 + z^2 = y^2$

By the converse of the Pythagorean theorem, $\Delta XYZ$ is a right-angled triangle with the right angle at vertex $Y$, i.e., $Y = 90^\circ = \frac{\pi}{2}$.

Now, let's analyze the given options:

1. Option (B): Since $X + Y + Z = 180^\circ$ and $Y = 90^\circ$, we have: $X + Z = 90^\circ = Y \implies Y = X + Z$ Thus, Option (B) is TRUE.

2. Option (C): In a right-angled triangle with $Y = 90^\circ$, we have: $\sin X = \frac{x}{y} \quad \text{and} \quad \cos X = \frac{z}{y}$

Using the half-angle identity $\tan\frac{X}{2} = \frac{\sin X}{1 + \cos X}$:

$\tan\frac{X}{2} = \frac{\frac{x}{y}}{1 + \frac{z}{y}} = \frac{x}{y + z}$ Thus, Option (C) is TRUE.

1. Option (A): Since $Y = 90^\circ$ and $X + Z = 90^\circ$, $2Y = 180^\circ \neq X + Z$. Thus, Option (A) is FALSE.

2. Option (D): Since $x^2 + z^2 = y^2$, we have $x^2 + z^2 - y^2 = 0 \neq xz$. Thus, Option (D) is FALSE.

Correct Options: (B) and (C)